1/-2-2 which is 1/-4 No? Obviously this is not equal to 1/0 so it is a defined value. Real numbers are all positive numbers including zero right?

Please elaborate here as how you got 1/0 by putting x = -2

b) Okay, I know that why we can't put 1 or 2 in here as it will make the fraction 1/0 which is undefined. But again my question why can't we use negative values for x here e.g If i put x = -2 doesn't that make the whole fraction

1/(-2-1)(-2-2) which is 1/(-3)(-4) = 1/12 isn't this a defined value?? Then why can't we use the negative values of x here?

Q2

I understand the working of this type of question but I don't understand why do we have to start the input value of x at 1 and why we can't use x = 0 in this equation? Isn't zero a positive real number?

If I put x = 0 then

f(0) = (0-1)^2 + 2 = 3 so my answer is f(x) is greater than or equal to 3

What am I doing wrong? When you say "The least that term can be is therefore 0", why?

Similarly I don't understand what you are trying to say in Q3.

Please elaborate how you are supposed to tackle these kind of questions. Maybe the method I'm doing is wrong.

1/-2-2 which is 1/-4 No? Obviously this is not equal to 1/0 so it is a defined value. Real numbers are all positive numbers including zero right?

Please elaborate here as how you got 1/0 by putting x = -2 Mr F says: It should be obvious that Lujan made a typo. S/he meant x = 2.

b) Okay, I know that why we can't put 1 or 2 in here as it will make the fraction 1/0 which is undefined. But again my question why can't we use negative values for x here e.g If i put x = -2 doesn't that make the whole fraction

1/(-2-1)(-2-2) which is 1/(-3)(-4) = 1/12 isn't this a defined value?? Then why can't we use the negative values of x here? Mr F says: You can. Who said you can't? x = 1 and x = 2 are the only values that are not allowed.

Q2

I understand the working of this type of question but I don't understand why do we have to start the input value of x at 1 and why we can't use x = 0 in this equation? Isn't zero a positive real number? Mr F says: No it's not. However, you can consider the limit of the function as x --> 0.

If I put x = 0 then

f(0) = (0-1)^2 + 2 = 3 so my answer is f(x) is greater than or equal to 3

What am I doing wrong? When you say "The least that term can be is therefore 0", why? Mr F says: Is x = 1 an allowed value? What is f(1)? You should be familiar with the graph of a parabola. Draw a graph of y = f(x) over the stated domain of all positive real numbers. What do you see?

Similarly I don't understand what you are trying to say in Q3. Mr F says: Draw the graphs.

Please elaborate how you are supposed to tackle these kind of questions. Maybe the method I'm doing is wrong.

Thanks, much appreciate the help guys. I will work on the above solutions and get back to you.

Keep up the good work

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Edit

Mr Fan.

You said that 0 is not a positive real number real but it is a real number right? Just not positive?
Also what do you mean by "However, you can consider the limit of the function as x --> 0"

Lastly you said that one should draw the graphs of the functions to find their ranges, but we can't do that for each function as it would take too much time and would not be accurate. Please correct me if I am wrong.

Thanks, much appreciate the help guys. I will work on the above solutions and get back to you.

Keep up the good work

======================

Edit

Mr Fan.

You said that 0 is not a positive real number real but it is a real number right? Mr F says: Yes. Just not positive? Mr F says: Correct.

Also what do you mean by "However, you can consider the limit of the function as x --> 0" Mr F says: x = 0 is not in the domain but x can approach 0 as a limit. Don't worry about this as it will probably be more confusing than helpful. When drawing the graph, just calculate f(0) and then put an open circle at the point to indicate that it's not included.

Lastly you said that one should draw the graphs of the functions to find their ranges, but we can't do that for each function as it would take too much time and would not be accurate. Please correct me if I am wrong. Mr F says: It shouldn't take very long to draw a parabola that is written in turning point form. You only want to use the graph to get the range, so you don't have to worry about things like x-intercepts. ONly show that informtaion that is relevant to finding the range. As I said, you should be able to do this in short order (if not, you need to start practicing).

This means that according to the traditional definition 0 is a positive integer. I am pretty sure I heard my A Level Maths teacher say that 0 is a positive integer.

Sir that's exactly what I lack, I need practice on sketching quadratic.

Could you please point me to an easy to follow guide or give me a few tips on how to sketch graphs especially of the following types:

Quadratic Graphs

I am familiar with the rules of parabolas (the graphs of y = ax^2+bx+c and what these variables mean). I just get confused when I see factorized forms. Please tell me how to make these graphs without taking a lot of time. I am willing to absorb and practice what ever you throw at me

I am familiar with the rules of parabolas (the graphs of y = ax^2+bx+c and what these variables mean). I just get confused when I see factorized forms. Please tell me how to make these graphs without taking a lot of time. I am willing to absorb and practice what ever you throw at me

Thanks in advance!

You went from a quadratic in step one to a cubic in step two...

To factorise think which numbers add to 1 and multiply to -4. Although don't think for too long because there are none. Use the quadratic formula to find x:

To plot the graph find a rough decimal approximation and plot it as , and . You can either find more pairs or join up using the correct shape.

Better yet, use a plotting program or a spreadsheet program to get a graph done (see attachment for